How to evaluate $\int _0^1\frac{\text{Li}_3\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x$

Question

This problem featuring the trilogarithm was posted online $$\int _0^1\frac{\text{Li}_3\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x$$ they did not leave a result and I wonder if a closed form exists.

What I tried in order to evaluate it was using that $$\int _0^1\frac{\text{Li}_3\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x=\sum _{n=1}^{\infty }\frac{1}{n^3}\int _0^1\frac{x^{n-1}}{\sqrt{1-x}}\:\mathrm{d}x$$ $$=2\sum _{n=1}^{\infty }\frac{1}{n^3}\int _0^{\frac{\pi }{2}}\sin \left(x\right)^{2n-1}\:\mathrm{d}x=2\sum _{n=1}^{\infty }\frac{\left(2n-2\right)!!}{n^3\left(2n-1\right)!!}$$ I am not sure how to proceed further with this method. Also $$\int _0^1\frac{\text{Li}_3\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x=2\int _0^1\frac{\text{arctanh} \left(\sqrt{1-x}\right)\text{Li}_2\left(x\right)}{x}\:\mathrm{d}x$$ $$=-\int _0^1\frac{\log \left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)\text{Li}_2\left(x\right)}{x}\:\mathrm{d}x$$ but this doesn't seem promising either. How else can this problem be approached? Does a closed form exist? I'd appreciate any help. Thanks!

Answer 1

By the repeated integral of Polylogarithm that $$ \operatorname{Li_3}(x) = \int_{0}^{1} \frac{\operatorname{Li_2}(xr)}{r} \,\mathrm{d}r,\quad \operatorname{Li_2}(x) = -\int_{0}^{1} \frac{\ln(1-xt)}{t} \,\mathrm{d}t $$ Continue with my comment, we have $$ \begin{aligned} \int_{0}^{1} \frac{\operatorname{Li_3}(x)}{x\sqrt{1-x}} \,\mathrm{d}x & = -\int_{0}^{1}\int_{0}^{1}\left(\int_{0}^{1}\frac{\ln(1-rtx)}{rtx\sqrt{1-x}}\,\mathrm{d}x\right)\mathrm{d}r\mathrm{d}t \\ & = 2\int_{0}^{1}\int_{0}^{1} \frac{\arcsin^2(\sqrt{rt})}{rt}\,\mathrm{d}r\mathrm{d}t = 2\int_{0}^{1} \frac1{t}\left(\int_{0}^{1} \frac{\arcsin^2(\sqrt{rt})}{r}\,\mathrm{d}r\right)\mathrm{d}t \\ (\text{IBP})\;& = 2\int_{0}^{1} \frac1{\sqrt{t}} \left(\int_{0}^{1} \frac{\arcsin(\sqrt{rt})\ln r}{\sqrt{r(1-rt)}}\,\mathrm{d}r\right)\mathrm{d}t \\ & = 2\int_{0}^{1} \frac{\ln r}{\sqrt{r}} \left(\int_{0}^{1} \frac{\arcsin(\sqrt{rt})}{\sqrt{t(1-rt)}}\,\mathrm{d}t\right)\mathrm{d}r \\ & = 2\int_{0}^{1} \frac{\ln r}{\sqrt{r}} \left(\frac{\arcsin^2(\sqrt{r})}{\sqrt{r}}\right)\mathrm{d}r = 2\int_{0}^{1} \frac{\arcsin^2(\sqrt{r})\ln r}{r}\,\mathrm{d}r \\ (\text{IBP}, r=\sin^2\!u)\; & = \int_{0}^{1} \frac{\arcsin(\sqrt{r})\ln^2\!r}{\sqrt{r(1-r)}}\,\mathrm{d}r = 8\int_{0}^{\pi/2} u\ln^2(\sin u)\,\mathrm{d}u \end{aligned} $$ then you can continue with this post to find your anwser.

Supplement: If you want to continue with your approach, recall that $$ 2\arcsin^2(x) = \sum_{n=1}^{\infty} \frac{2^{2n}x^{2n}}{n^2\binom{2n}{n}} $$ note your series is $$ 2\sum_{n=1}^{\infty} \frac{(2n-2)!!}{n^3(2n-1)!!} = \sum_{n=1}^{\infty} \frac{2^{2n}}{n^4\binom{2n}{n}} = 8\int_{0}^{1} \frac1{t}\left(\int_{0}^{t} \frac{\arcsin^2(r)}{r}\,\mathrm{d}r\right)\mathrm{d}t $$ which is exactly the same integral from above.

Answer 2

Using, as you wrote, $$I_3=\int _0^1\frac{\text{Li}_3\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x=\sum _{n=1}^{\infty }\frac{1}{n^3}\int _0^1\frac{x^{n-1}}{\sqrt{1-x}}\:\mathrm{d}x$$ $$I_3=\sum _{n=1}^{\infty }\frac{\sqrt \pi}{n^3}\frac{\Gamma (n)}{\Gamma \left(n+\frac{1}{2}\right)}=2 \,\, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;1\right)$$ which can be generalized to $$I_k=\int _0^1\frac{\text{Li}_k\left(x\right)}{x\sqrt{1-x}}\:\mathrm{d}x$$ $$I_4=2 \,\, _6F_5\left(1,1,1,1,1,1;\frac{3}{2},2,2,2,2;1\right)$$ $$I_5=2 \, _7F_6\left(1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2;1\right)$$ and you see the pattern.

Beside the hypergeometric functions, I do not see any possible result.

Answer 3

Too long to be a comment.

Let $x = \cos \theta$, then $dx = -\sin \theta . d\theta$

The integral becomes $$ \int_{0}^{\pi/2}\dfrac{\operatorname{Li}_3(\cos \theta)\sin\theta}{\cos\theta\sqrt{1-\cos\theta}}.d\theta\\ =\sqrt{2} \int_{0}^{\pi/2}\dfrac{\operatorname{\operatorname{Li_3(\cos\theta)\cos(\theta/2)}}}{\cos\theta}.d\theta = \dfrac{1}{\sqrt{2}}\int_{0}^{\pi/2}\dfrac{\operatorname{Li_3}(\sin\theta)}{\sin(\theta/2)}.d\theta \approx 2.237 $$

I don't think any simple closed form is possible